Every $n\times n$ matrix is the sum of a diagonalizable matrix and a nilpotent matrix.

Question

I would like to prove that every $n\times n$ matrix is the sum of a diagonalizable matrix and a nilpotent matrix. How is this possible? I'm not sure where to begin really- I know that a nilpotent matrix is one of which some power is the zero matrix. I also know that a matrix A can be written as $AP=PJ$ with $P$ invertible and $J$ of Jordan form. I have proven that any strictly upper triangular matrix is nilpotent, so $J$ can be written as $D+N $, with D diagonal and $N$ nilpotent, but how can I change this for A? Thank you!

Answer 1

You have $A = PJP^{-1}$ where $J$ is in Jordan form. Write $J = D + N$ where $D$ is the diagonal and $N$ is the rest, which is strictly upper triangular and thus nilpotent. Then $A = PDP^{-1} + PNP^{-1}$. The former is clearly diagonalizable, while the latter is nilpotent; just note that $(PNP^{-1})(PNP^{-1}) = PN(P^{-1}P)NP^{-1} = PN^2P^{-1}$ and so on.

Answer 2

This is known as the Jordan--Chevalley decomposition of an operator, and its proof is very elementary. In particular, you do not need the Jordan normal form here.

Let $T: V \longrightarrow V$ be any endomorphism. We know that $T$ admits finitely many eigenvalues $\lambda$ with certain multiplicities $m(\lambda)$, and that $V$ decomposes into the generalized eigenspaces $V_\lambda = \ker(T-\lambda)^{m(\lambda)}$. Consider now the operator $T_s$ which acts on $V_\lambda$ through multiplication by $\lambda$. Thus $T_s$ corresponds to a diagonal matrix with $\lambda$ appearing as many times as the dimension of $V_\lambda$.

Notice that now $T_n = T-T_s$ is nilpotent on each $V_\lambda$ and, since there are finitely many such summands, $T_n$ is nilpotent on all of $V$. Finally, $T$ commutes with $T_s$, which you can check on each component, so it also commutes with $T_n$, which is what was required.